Final answer:
To find the 90th percentile of a normal population with mean 60 and standard deviation 9, calculate the z-score for the 90th percentile, multiply by the standard deviation, and add to the mean, resulting in approximately 71.52.
Step-by-step explanation:
To determine the 90th percentile of a normal distribution with a given mean (μ = 60) and standard deviation (σ = 9), we would use the z-score corresponding to the 90th percentile in a standard normal distribution and then translate that back to our specific distribution. Calculating the 90th percentile in a standard normal distribution typically involves using a z-table or statistical software, but here's the general process:
- Find the z-score that corresponds to the 90th percentile, which is typically about 1.28.
- Multiply this z-score by the standard deviation of the population. In this case, 1.28 * 9 = 11.52.
- Add the product to the mean of the distribution to get the 90th percentile value. So, 60 + 11.52 = 71.52.
Therefore, the 90th percentile of the population would be approximately 71.52.